What's the rationale behind rejecting limits of recurrence relation?
Yes, we can reject limit for other reason. After all, the relation $L=1/(4-L)$ is merely a necessary condition, not a sufficient one.
It is plausible that the limit is actually $3.732\dotso$ instead, had you pick a different starting point (for example, if you start at $3.732\dotso$).
These limit are fixed points of the recurrent relation. Some fixed point are attractive, and some are not, so it depends on the situation.
For this kind of recurrent relation, a well-known method is known to solve exactly for the sequence. Consider $y_{i}=x_{i}-L$. Then \begin{align} &y_{i+1}=x_{i+1}-L=\frac{1}{4-x_{i}}-L=\frac{1-4L+Lx_{i}}{4-x_{i}}=\frac{1-4L+L^{2}+Ly_{i}}{4-L-y_{i}}=\frac{Ly_{i}}{4-L-y_{i}} \\ &\qquad\qquad\implies \frac{1}{y_{i+1}}=\frac{4-L-y_{i}}{Ly_{i}}=\left(\frac{4}{L}-1\right)\frac{1}{y_{i}}-\frac{1}{L} \\ &\qquad\qquad\implies \frac{1}{y_{i+1}}-\left(\frac{4}{L}-1\right)\frac{1}{y_{i}}=-\frac{1}{L}. \end{align} Write $z_{i}=\frac{1}{y_{i}}$ and we have $$z_{i+1}-\left(\frac{4}{L}-1\right)z_{i}=-\frac{1}{L}.$$ This is now a standard first order inhomogeneous linear recurrent relation and can be solved exactly, and from there you can figure out precisely whether the original sequence converge to $L$ or not by checking whether $|z_{i}|$ go to infinity.
EDIT: let me add in that it's easy to see from the recurrent relation for $z_{i}$ exactly how the sequence behave in the limit without solving exactly for $\{z_{i}\}$. If you pick $L=3.732\dotso$ then $\frac{4}{L}-1<1$ so it is impossible for $|z_{i}|$ to go to infinity, because when $|z_{i}|$ is sufficiently large then $$|z_{i+1}|=\left|\left(\frac{4}{L}-1\right)z_{i}-\frac{1}{L}\right|<|z_{i}|.$$ So it's impossible for any starting point other than $3.732\dotso$ itself to reach there (the argument doesn't apply for that starting point because then $z_{i}$ are undefined), this fixed point is unstable. But for $L=0.268\dotso$ then $\frac{4}{L}-1$ is large, so for almost all values of $z_{i}$ it should blow up.
A way to reason could be this. First you prove some kind of monotony, by induction. For instance, your sequence looks decreasing to me, Then prove that it is limited from below, that is it can’t go to $- \infty$ (easy to check in your case).
You know that every monotone sequence admits a limit, which can be infinite, but we’ve ruled out that possibility. Call that limit $l$ and for $n \to \infty$ you have $l=\frac{1}{4-l}$. Note that here you are using the continuity of the function $\frac{1}{4-x}$ in the point $l$, since it is discontinuous only in $4$ and you know $l \leq 3$. Now find $l$, and as you can see by yourself there is only one admissible solution, and 3,732 can’t be the answer because of the fact the sequence is decreasing and starting from 3.
A good exercise would be to study how are these “equilibrium points”. I mean, if you start at $l$ equal the other solution, it is obvious you have a fixed point, but... what about points near it? If you start from there, what happens, where do you end up? Try to think about this
Suggestion: Start using technology to actually compute some values for problems like this, and using that to build up intuition about what the process or sequence is doing. Admittedly this is not a core traditional technique in analysis, but I always worked this way through my schooling, and I felt like it put me a step or two ahead of my classmates. Of course it's not a deductive proof, but the intuition should put you on track to answering a number of questions like you posed here yourself.
This could be done with a simple program or a spreadsheet. In this case, the following took about 20 seconds to generate in an open-source spreadsheet:
